Checking if a string contains all blankspace in R

Question

Im looking to see if a string contains only blank space. The string could be

"  "

or

"           "

or

"              " 

etc...

I want to do this so I can change values in a data frame to NA, because my goal is to fix/clean messed up data.

Thank you

Solution 1

You can try with grepl:

grepl("^\\s*$", your_string)

"^\\s*$" asks for 0 or more (*) spaces (\\s) between beginning (^) and end ($) of string.

Examples

grepl("^\\s*$", " ")
#[1] TRUE
grepl("^\\s*$", "")
#[1] TRUE
grepl("^\\s*$", "    ")
#[1] TRUE
grepl("^\\s*$", " ab")
[1] FALSE

NB: you can also just use a space instead of \\s in the regex ("^\\s*$").

Solution 2

Without regex, you could use

which(nchar(trimws(vec))==0)

The function trimws() removes trailing and leading whitespace characters from a string. Hence, if after the use of trimws the length of the string (determined by nchar()) is not zero, the string contains at least one non-whitespace character.

Example:

vec <- c(" ", "", "   "," a  ", "             ", "b")
which(nchar(trimws(vec))==0)
#[1] 1 2 3 5

The entries 1, 2, 3, and 5 of the vector vec are either empty or contain only whitespace characters.

---

As suggested by Richard Scriven, the same result can be obtained without calling nchar(), by simply using trimws(vec)=="" (or which(trimws(vec)==""), depending on the desired output: the former results in a vector of booleans, the latter in the index numbers of the blank/empty entries).

Comments

[Seekheart]

a better regex would be grepl("\\s+", your_string, perl = True) the + forces it to match 1 more spaces.

[Cath]

@Seekheart it is actually why I prefered * so it also checks for the empty string, I thought the OP might also want empty strings to be spotted (and further modified as NA) (btw, yours wouldn't work without specifying beginning and end of string, you don't need perl option for that and TRUE must be written in uppercase ;-) )